Question

find a point on the y axis egudistant from (-5,2)and (9,-2)

Answer 1

$\text{Answer : } \\ \\ \text{Let, the point on y - axis be (0, y)}\\ \\ \text {Then, the distance between the points } \\ \text{(0, y) and ( - 5, 2) is} \\ = \sqrt{ { \{0 - ( -5 ) \}}^{2} + {( \text{y} - 2)}^{2} } \\ = \sqrt{25 + { \text{y}}^{2} - \text{4y} + 4} \\ = \sqrt{ { \text{y}}^{2} - \text{4y} + 29 } \: \: \text{units}\\ \\ \text{and the distance between the points} \\ \text{(0, y) and (9, - 2) is} \\ = \sqrt{ {(0 - 9)}^{2} + { \{ \text{y} -( - 2) \} }^{2} } \\ = \sqrt{81 + { \text{y}}^{2} + \text{4y} + 4 } \\ = \sqrt{ { \text{y}}^{2} + \text{4y} + 85 } \: \: \text{units} \\ \\ \text{By the given condition,} \\ \sqrt{ { \text{y}}^{2} - \text{4y} + 29 } = \sqrt{ { \text{y}}^{2} + \text{4y} + 85 } \\ \\ \text{Squaring both sides, we get} \\ { \text{y}}^{2} - \text{4y} + 29 = { \text{y}}^{2} + \text{4y} + 85 \\ \to \text{4y + 4y = 29 - 85} \\ \to \text{8y = - 56} \\ \to \text{y} = - \frac{56}{8} \\ \therefore \text{y = - 7 } \\ \\ \therefore \text{The required point on y - axis is (0, - 7).}$

Answer 2

Hey mate !

Let the point on y axis is A be (0,y)

Now the condition is given that y axis is equidistant

so the

( 0,y )distance from ( -5,2 ) = ( 0,y) distance from ( 9,-2)

We have a formula
$\sqrt{ ( x_{2} - x_{1}) {}^{2} + ( y_{2} - y_{1}} ) {}^{2}$

so

√ is also written as 1/2

((0-(-5))² +(y-2)²)^1/2 = ((0-(9)²) + (y-(-2))²)^1/2

Now squaring both side

{(5)²+(y²+4-4y)}^1/2×2 = {(-9)²+(y²+4+4y)}^1/2×2

{25+y²+4-4y} = { 81 + y²+4+4y}

25-4y = 81 +4y

-4y-4y = 81 -25

-8y = 56

y = -56/8

y = -7

answer

[the point on y - axis is become (0,y) = (0,-7) ]